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A satellite is orbiting close to the earth and has a kinetic energy \( \mathrm{K} \). The minimum extra kinetic
energy required by it to just overcome the gravitation pull of the earth is
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energy required by it to just overcome the gravitation pull of the earth is
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The correct answer is:
\( \mathrm{K} \)
(B)
We know \( v_{e}=\sqrt{2} v_{0} \)
When satellite is in orbit \( K=\frac{1}{2} m v_{0}{ }^{2} \)
For satellite to escape its velocity should be \( v_{e} \)
\( \therefore K E \) To escape, \( K_{e}=\frac{1}{2} m v_{e}^{2}=\frac{1}{2} m\left(\sqrt{2} v_{0}\right)^{2}=2 K \)
\( \therefore \) Extra \( K . E \). required \( =2 K-K=K \)
We know \( v_{e}=\sqrt{2} v_{0} \)
When satellite is in orbit \( K=\frac{1}{2} m v_{0}{ }^{2} \)
For satellite to escape its velocity should be \( v_{e} \)
\( \therefore K E \) To escape, \( K_{e}=\frac{1}{2} m v_{e}^{2}=\frac{1}{2} m\left(\sqrt{2} v_{0}\right)^{2}=2 K \)
\( \therefore \) Extra \( K . E \). required \( =2 K-K=K \)
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